HDUOJ_2700_Parity

最新推荐文章于 2017-12-18 20:34:20 发布

momo-R

最新推荐文章于 2017-12-18 20:34:20 发布

阅读量223

点赞数 1

CC 4.0 BY-SA版权

分类专栏： HDUOJ

本文链接：https://blog.youkuaiyun.com/stanly_shen/article/details/51231815

HDUOJ 专栏收录该内容

7 篇文章

订阅专栏

Parity

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4020 Accepted Submission(s): 3020

Problem Description

A bit string has odd parity if the number of 1's is odd. A bit string has even parity if the number of 1's is even.Zero is considered to be an even number, so a bit string with no 1's has even parity. Note that the number of
0's does not affect the parity of a bit string.

Input

The input consists of one or more strings, each on a line by itself, followed by a line containing only "#" that signals the end of the input. Each string contains 1–31 bits followed by either a lowercase letter 'e' or a lowercase letter 'o'.

Output

Each line of output must look just like the corresponding line of input, except that the letter at the end is replaced by the correct bit so that the entire bit string has even parity (if the letter was 'e') or odd parity (if the letter was 'o').

Sample Input

Sample Output

Source

2008 Mid-Central USA

#include <stdio.h>
#include <string.h>
int main(){
    char a[101];
    while (gets(a)&&strcmp(a,"#")) {
        int i;
        int n0=0,n1=0;
        int len=(int)strlen(a);
        for (i=0; i<len; i++) {
            if (a[i]=='0') {
                n0++;
            }
            else if (a[i]=='1'){
                n1++;
            }
        }
        if (a[len-1]=='e') {
            if (n1%2==0) {
                a[len-1]='0';
            }
            else
                a[len-1]='1';
        }
        else if (a[len-1]=='o'){
            if (n1%2==0) {
                a[len-1]='1';
            }
            else
                a[len-1]='0';
        }
        puts(a);
    }
    return 0;
}